Answer:
Explanation:
Given that,
Mass of oxygen M = 4kg
Temperature of oxygen T.i = 20°C
T.i = 20+273 =293K
Specific heat capacity of oxygen
C = 0.658 J/kgK
And at pressure P = 150Kpa
Mass of Nitrogen m = 7kg
Temperature of Nitrogen t.i = 40°C
t.i = 40+273 = 313K
And at pressure p = 100Kpa
Specific heat capacity of nitrogen
c = 0.743 J/kgK
a. Mixture temperature can be determine from the energy balance of the container
Using Heat energy formula
∆U = 0
For oxygen
H.o = MC∆θ
H.o = MC(T.f - T.i)
H.o = 4×0.658 (T.f - 293)
H.o = 2.632(T.f - 293)
For Nitrogen
H.n = mc∆θ
H.n = mc(T.f - t.i)
H.n = 7×0.743(T.f-313)
H.n = 5.201(T.f-313)
Then,
∆U = 0
H.o + H.n = 0
2.632(T.f - 293) + 5.201(T.f-313) = 0
2.632T.f -771.176+ 5.201T.f-1627.9=0
7.833T.f - 2399.089 = 0
7.833T.f = 2399.089
T.f = 2399.089/7.833
T.f = 306.28 K
b. Number of molecules n?
Number of mole can be calculated by using the formula
n = mass/molar mass
For oxygen,
Molar mass of oxygen O2
O2 = 16×2 = 32g/mol
n = mass of oxygen / molar mass of oxygen
n = 4000/32
n = 125 mole
For Nitrogen
Molar mass of nitrogen N.2
N.2 = 14×2 = 28g/mol
n = mass of nitrogen /molar mass of nitrogen
n' = 7000/28
n' = 250 mole
C. The initial volume of each gas V?
Volume can be calculated using the ideal gas law
PV = nRT
V = nRT/P
R is a constant = 8.314 J/mol•K
For nitrogen
V = n'Rt/p
V = 250 × 8.314 ×313 / 100,000
Vn = 6.51m³
For Oxygen
V = nRT/P
V = 125 × 8.314 × 293 / 150,000
Vo = 2.03m³
d. Final pressure mixture
Molar mass of mixture
M = total mass/total mole
M = (7000+4000)/(250+125)
M = 29.333g/mol
The mixture total number of mole
N = mole of oxygen + mole of nitrogen
N = 250+125 = 375 mol
Then, using ideal gas law
PV = nRT
P = nRT/V
Total mole = 375mol
R = 8.314 J/mol•K
Mixture Temperature T.f = 306.28K
Total Volume Vo+Vn = 2.03+6.51
V = 8.54m³
Then,
P = nRT/V
P = 375×8.314×306.28/8.54
P = 111,815.5 Pa
To K•pa
P = 111.82 K•Pa
